l1.next=merge(l1.next, l2);returnl1; }else{ l2.next=merge(l1, l2.next);returnl2; } } } 三刷: 依然是使用之前的方法,找中点,再递归进行mergesort。 Java: Time Complexity - O(nlogn), Space Complexity - O(logn)。 /*** Definition for
Sort a linked list inO(nlogn) time using constant space complexity. 归并排序(Merge Sort,台湾译作:合并排序)是建立在归并操作上的一种有效的排序算法。该算法是采用分治法(Divide and Conquer)的一个非常典型的应用。 归并操作(Merge),也叫归并算法,指的是将两个已经排序的序列合并成一个序列的操作。归并排...
描述 Sort a linked list inO(n log n)time using constant space complexity. 分析 常数空间且O(nlogn),单链表适合用归并排序,双向链表适合用快速排序。本题可以复用Merge Two Sorted Lists的代码。 代码 // Sort List// 归并排序,时间复杂度O(nlogn),空间复杂度O(1)publicclassSolution{publicListNodesortLi...
linked list head pointer, compute and return the number of nodes in the list. */intLength(list_t* list)//node 1 is 1{ printf("in length\n"); element_t* current = list->head;intcount = 0;while(current != NULL) { printf("in length while\n"); count++; current = current->...
Sort a linked list in O(n log n) time using constant space complexity. 本题就是考察的是链表的归并排序。 代码如下: /*class ListNode { int val; ListNode next; ListNode(int x) { val = x; } }*/ public class Solution { public ListNode sortList(ListNode head) ...
Sort a linked list using insertion sort. A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list. With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list ...
Here are some of the applications of merge sort which are explained below: Sorting linked lists:Since random access of elements takes much time in case of the linked list, Merge Sort provides a quick solution to sort the linked list elements. This also stores the elements in self-made arrays...
C++ Exercises, Practice and Solution: Write a C++ program to sort the elements of a stack (using a linked list).
linked list, using constant space complexity. """ def sortList(self, head): # write your code here def merge(list1,list2): if list1 == None: return list2 if list2 == None: return list1 head = None if list1.val < list2.val: head = list1 list1 = list1.next else: head =...
linked list, using constant space complexity. """ def sortList(self, head): # write your code here def merge(list1,list2): if list1 == None: return list2 if list2 == None: return list1 head = None if list1.val < list2.val: head = list1 list1 = list1.next else: head =...